Answer:
x = 3
, y = 5
Step-by-step explanation:
Solve the following system:
{3 y - 7 x = -6 | (equation 1)
3 y - 3 x = 6 | (equation 2)
Subtract 3/7 × (equation 1) from equation 2:
{-(7 x) + 3 y = -6 | (equation 1)
0 x+(12 y)/7 = 60/7 | (equation 2)
Multiply equation 2 by 7/12:
{-(7 x) + 3 y = -6 | (equation 1)
0 x+y = 5 | (equation 2)
Subtract 3 × (equation 2) from equation 1:
{-(7 x)+0 y = -21 | (equation 1)
0 x+y = 5 | (equation 2)
Divide equation 1 by -7:
{x+0 y = 3 | (equation 1)
0 x+y = 5 | (equation 2)
Collect results:
Answer: {x = 3
, y = 5
Answer:
A quadratic equation can be written as:
a*x^2 + b*x + c = 0.
where a, b and c are real numbers.
The solutions of this equation can be found by the equation:
Where the determinant is D = b^2 - 4*a*c.
Now, if D>0
we have the square root of a positive number, which will be equal to a real number.
√D = R
then the solutions are:
Where each sign of R is a different solution for the equation.
If D< 0, we have the square root of a negative number, then we have a complex component:
√D = i*R
We have two complex solutions.
If D = 0
√0 = 0
then:
We have only one real solution (or two equal solutions, depending on how you see it)